HDU2126 扩展背包 DP

Buy the souvenirs

Time Limit: 10000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 260    Accepted Submission(s): 69

Problem Description

When the winter holiday comes, a lot of people will have a trip. Generally, there are a lot of souvenirs to sell, and sometimes the travelers will buy some ones with pleasure. Not only can they give the souvenirs to their friends and families as gifts, but also can the souvenirs leave them good recollections. All in all, the prices of souvenirs are not very dear, and the souvenirs are also very lovable and interesting. But the money the people have is under the control. They can’t buy a lot, but only a few. So after they admire all the souvenirs, they decide to buy some ones, and they have many combinations to select, but there are no two ones with the same kind in any combination. Now there is a blank written by the names and prices of the souvenirs, as a top coder all around the world, you should calculate how many selections you have, and any selection owns the most kinds of different souvenirs. For instance:

 

And you have only 7 RMB, this time you can select any combination with 3 kinds of souvenirs at most, so the selections of 3 kinds of souvenirs are ABC (6), ABD (7). But if you have 8 RMB, the selections with the most kinds of souvenirs are ABC (6), ABD (7), ACD (8), and if you have 10 RMB, there is only one selection with the most kinds of souvenirs to you: ABCD (10).

 

 

Input

For the first line, there is a T means the number cases, then T cases follow.
In each case, in the first line there are two integer n and m, n is the number of the souvenirs and m is the money you have. The second line contains n integers; each integer describes a kind of souvenir.
All the numbers and results are in the range of 32-signed integer, and 0<=m<=500, 0<n<=30, t<=500, and the prices are all positive integers. There is a blank line between two cases.

 

 

Output

If you can buy some souvenirs, you should print the result with the same formation as “You have S selection(s) to buy with K kind(s) of souvenirs”, where the K means the most kinds of souvenirs you can buy, and S means the numbers of the combinations you can buy with the K kinds of souvenirs combination. But sometimes you can buy nothing, so you must print the result “Sorry, you can't buy anything.”

 

 

Sample Input

2

4 7

1 2 3 4

 

4 0

1 2 3 4

 

 

Sample Output

You have 2 selection(s) to buy with 3 kind(s) of souvenirs.

Sorry, you can't buy anything.

 

 

 

 

背包的一种进化版,除了记录最多能买多少个,需要记录买这么多个的方法,所以要在二维的基础上加多一维.

状态转移方程如下: f[i][j][k]=f[i-1][j-t[i]][k-1]+f[i-1][j][k];

(前i个物品在有j元的时候买k个物品的方法,t[i]为第i个物品的价格)

 

 

代码如下:

#include<iostream>using namespace std;int i,j,m,n,p,t[31],k,maxxx,a,b,s,g;int f[31][501][431];int main(){scanf("%d",&p);for (g=1;g<=p;++g){scanf("%d%d",&n,&m);maxxx=0;for (i=1;i<=n;++i) scanf("%d",&t[i]);for (i=1;i<=n;++i)for (j=1;j<=m;++j)for (k=1;k<=i;++k)if (j>=t[i]){if (k==1) f[i][j][k]=f[i-1][j][k]+1;//由于k==0时没有进行初始化,所以当k==1时需要另外处理else f[i][j][k]+=(f[i-1][j-t[i]][k-1]+f[i-1][j][k]);}else f[i][j][k]=f[i-1][j][k];//当第i个物品买不了的时候,f[i][j][k]的值就是f[i-1][j][k]的值for (i=1;i<=n;++i) if (f[n][m][i]!=0) maxxx=i;//求最大购买物品数maxxxif (maxxx!=0) printf("You have %d selection(s) to buy with %d kind(s) of souvenirs./n",f[n][m][maxxx],maxxx);else printf("Sorry, you can't buy anything./n");}return 0;}

 

 

 

此算法的时间复杂度为O(n*n*m),虽然复杂度不高,不过无论是时间还是空间还是有很大的优化余地.

 

另一种算法(O(n*m)):

在二维背包的基础上加多一维只有0和1两个下标的第三维,其中f[i][j][0]表示j元在前i个物品中最多能买多少个物品,f[i][j][1]表示买最多物品的方法数.

状态转移方程为:(v[i]为第i个物品的价格)

f[i][j][0]=max(f[i-1][j][0],f[i-1][j-v[i]][0]+1);

if  (f[i-1][j][0]==f[i-1][j-v[i]][0]) f[i][j][1]= f[i-1][j-v[i]][1]+f[i-1][j][1];

else  if  (f[i-1][j][0]>f[i-1][j-v[i]][0]) f[i][j][1]=f[i-1][j][0];

else  f[i][j][1]= f[i-1][j-v[i]][1];

 

 

代码如下:

#include<iostream>using namespace std;int p,n,m,i,j,f[31][501][2],v[31];int main(){scanf("%d",&p);while (p--){scanf("%d%d",&n,&m);for (i=1;i<=n;++i) scanf("%d",&v[i]);for (i=0;i<=n;++i) f[i][0][1]=1;for (j=0;j<=m;++j) f[0][j][1]=1;for (i=1;i<=n;++i)for (j=1;j<=m;++j)if (j>=v[i]){if (f[i-1][j-v[i]][0]+1>f[i-1][j][0]){f[i][j][0]=f[i-1][j-v[i]][0]+1;f[i][j][1]=f[i-1][j-v[i]][1];}else if (f[i-1][j-v[i]][0]+1<f[i-1][j][0]){f[i][j][0]=f[i-1][j][0];f[i][j][1]=f[i-1][j][1];}else {f[i][j][0]=f[i-1][j-v[i]][0]+1;f[i][j][1]=f[i-1][j-v[i]][1]+f[i-1][j][1];}}else {f[i][j][0]=f[i-1][j][0];f[i][j][1]=f[i-1][j][1];}if (f[n][m][0]) printf("You have %d selection(s) to buy with %d kind(s) of souvenirs./n",f[n][m][1],f[n][m][0]);else printf("Sorry, you can't buy anything./n");}return 0;}